C++程序设计实践指导1.5求两个整数集合并集改写要求实现

改写要求1：改写为单链表结构可以对任意长度整数集合求并集

#include <cstdlib>

#include <iostream>



using namespace std;

struct LinkNode

{

  int data;

  LinkNode* next;     

};

class SET

{

      public:

             struct LinkNode* creat(int x[],int len);

             struct LinkNode* copy(LinkNode* aHead);

             int notin(int elem,LinkNode* cHead)

             {

                 LinkNode* p;

                 p=cHead;

                 while(p)

                 {

                         if(elem==p->data)

                         return 0;

                         p=p->next;   

                 }       

                 return 1;    

             }

             

             void insert(LinkNode* bHead,LinkNode* cHead);

             void output(LinkNode* cHead)

             {

                  LinkNode* p=cHead->next;

                  while(p)

                  {

                          cout<<p->data<<"\t";

                          p=p->next;

                  }

                  cout<<endl;

             }

};



struct LinkNode* SET::creat(int x[],int len)

{

       LinkNode* pHead=new LinkNode;

       pHead->next=NULL;

       LinkNode* p;

       p=pHead;

       for(int i=0;i<len;i++)

       {

             LinkNode* newLinkNode=new LinkNode;

             newLinkNode->next=NULL;

             newLinkNode->data=x[i];

             p->next=newLinkNode;

             p=newLinkNode;

       }

       return pHead;

}



struct LinkNode* SET::copy(LinkNode* aHead)

{

     LinkNode* cHead=new LinkNode;

     cHead->next=NULL;  

     LinkNode* p,*r;

     p=aHead->next;

     r=cHead;

     while(p)

     {

             LinkNode* newLinkNode=new LinkNode;

             newLinkNode->next=NULL;

             newLinkNode->data=p->data;

             p=p->next;

             r->next=newLinkNode;

             r=newLinkNode;

     }

     return cHead;

}



void SET::insert(LinkNode *bHead,LinkNode* cHead)

{

     

     LinkNode* q,*s,*t;

     q=bHead->next;    

     s=cHead->next;

     while(s)

     {

             t=s;

             s=s->next;

     }

    while(q)

     {

             if(notin(q->data,cHead)!=0)

             {

                 LinkNode* newLinkNode=new LinkNode;

                 newLinkNode->next=NULL;

                 newLinkNode->data=q->data;  

                 t->next=newLinkNode;

                 t=newLinkNode;

             }

             q=q->next; 

     }

    

}



int main(int argc, char *argv[])

{

    int s1[]={1,3,5,7,9};

    int s2[]={1,2,3,4,5,6};

    LinkNode* aHead,*bHead,*cHead,*Head;

    SET set;

    aHead=set.creat(s1,sizeof(s1)/sizeof(s1[0]));

    bHead=set.creat(s2,sizeof(s2)/sizeof(s2[0]));    

    cHead=set.copy(aHead);

    set.insert(bHead,cHead);

    set.output(cHead);

    system("PAUSE");

    return EXIT_SUCCESS;

}

改写要求2：对任意长度的两个整数集合求交集

#include <cstdlib>

#include <iostream>



using namespace std;

struct LinkNode

{

  int data;

  LinkNode* next;     

};

class SET

{

      public:

             struct LinkNode* creat(int x[],int len);

             struct LinkNode* copy(LinkNode* aHead);

             int notin(int elem,LinkNode* cHead)

             {

                 LinkNode* p;

                 p=cHead;

                 while(p)

                 {

                         if(elem==p->data)

                         return 1;

                         p=p->next;   

                 }       

                 return 0;    

             }

             

             struct LinkNode* insert(LinkNode* bHead,LinkNode* cHead);

             void output(LinkNode* cHead)

             {

                  LinkNode* p=cHead->next;

                  while(p)

                  {

                          cout<<p->data<<"\t";

                          p=p->next;

                  }

                  cout<<endl;

             }

};



struct LinkNode* SET::creat(int x[],int len)

{

       LinkNode* pHead=new LinkNode;

       pHead->next=NULL;

       LinkNode* p;

       p=pHead;

       for(int i=0;i<len;i++)

       {

             LinkNode* newLinkNode=new LinkNode;

             newLinkNode->next=NULL;

             newLinkNode->data=x[i];

             p->next=newLinkNode;

             p=newLinkNode;

       }

       return pHead;

}



struct LinkNode* SET::copy(LinkNode* aHead)

{

     LinkNode* cHead=new LinkNode;

     cHead->next=NULL;  

     LinkNode* p,*r;

     p=aHead->next;

     r=cHead;

     while(p)

     {

             LinkNode* newLinkNode=new LinkNode;

             newLinkNode->next=NULL;

             newLinkNode->data=p->data;

             p=p->next;

             r->next=newLinkNode;

             r=newLinkNode;

     }

     return cHead;

}



struct LinkNode* SET::insert(LinkNode *bHead,LinkNode* cHead)

{

     

     LinkNode* q,*t;

     LinkNode* Head=new LinkNode;

     q=bHead->next;

     Head->next=NULL;

     t=Head;    

    while(q)

     {

             if(notin(q->data,cHead)!=0)

             {

                 LinkNode* newLinkNode=new LinkNode;

                 newLinkNode->next=NULL;

                 newLinkNode->data=q->data;  

                 t->next=newLinkNode;

                 t=newLinkNode;

             }

             q=q->next; 

     }

     return Head;

}



int main(int argc, char *argv[])

{

    int s1[]={1,3,5,7,9};

    int s2[]={1,2,3,4,5,6};

    LinkNode* aHead,*bHead,*cHead,*Head;

    SET set;

    aHead=set.creat(s1,sizeof(s1)/sizeof(s1[0]));

    bHead=set.creat(s2,sizeof(s2)/sizeof(s2[0]));    

    cHead=set.copy(aHead);

    Head=set.insert(bHead,cHead);

    set.output(Head);

    system("PAUSE");

    return EXIT_SUCCESS;

}